# How do you integrate (x) / (x^(3)-x^(2)-2x +2) using partial fractions?

$= \setminus \ln | \setminus \frac{\setminus \sqrt{{x}^{2} - 2}}{x - 1} | + \frac{1}{\setminus \sqrt{2}} \setminus \ln | \setminus \frac{x - \setminus \sqrt{2}}{x + \setminus \sqrt{2}} | + C$

#### Explanation:

Making partial fractions as follows

$\setminus \frac{x}{{x}^{3} - {x}^{2} - 2 x + 2}$

$= \setminus \frac{x}{\left(x - 1\right) \left({x}^{2} - 2\right)}$

$= \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} - 2}$

Comparing corresponding coefficients on both sides we get

$A = - 1 , B = 1 , C = 2$

Now,

$\setminus \frac{x}{{x}^{3} - {x}^{2} - 2 x + 2} = - \frac{1}{x - 1} + \frac{x + 2}{{x}^{2} - 2}$

Integrating above equation on both the sides w.r.t. $x$, we get

$\setminus \int \setminus \frac{x}{{x}^{3} - {x}^{2} - 2 x + 2} \setminus \mathrm{dx} = \setminus \int \left(- \frac{1}{x - 1} + \frac{x + 2}{{x}^{2} - 2}\right) \setminus \mathrm{dx}$

$= - \setminus \int \frac{1}{x - 1} \setminus \mathrm{dx} + \setminus \int \frac{x}{{x}^{2} - 2} \setminus \mathrm{dx} + 2 \setminus \int \frac{1}{{x}^{2} - 2} \setminus \mathrm{dx}$

$= - \ln | x - 1 | + \frac{1}{2} \setminus \int \frac{d \left({x}^{2} - 2\right)}{{x}^{2} - 2} \setminus \mathrm{dx} + 2 \setminus \int \frac{1}{{x}^{2} - {\left(\setminus \sqrt{2}\right)}^{2}} \setminus \mathrm{dx}$

$= - \ln | x - 1 | + \frac{1}{2} \setminus \ln | {x}^{2} - 2 | + 2 \setminus \cdot \frac{1}{2 \setminus \sqrt{2}} \setminus \ln | \setminus \frac{x - \setminus \sqrt{2}}{x + \setminus \sqrt{2}} | + C$

$= \setminus \ln | \setminus \frac{\setminus \sqrt{{x}^{2} - 2}}{x - 1} | + \frac{1}{\setminus \sqrt{2}} \setminus \ln | \setminus \frac{x - \setminus \sqrt{2}}{x + \setminus \sqrt{2}} | + C$